Calculate the difference quotient Worksheet Name: Date: \( \qquad \) Period: \( \qquad \) 1. Find the difference quotient of the function below. \[ f(b)=-7 b-7 \] 2. Find the difference quotient of the function below. \[ f(k)=4 k-5 \] 4. Find the difference quotient of the function below. \[ f(c)=-2 c-9 \] 6. Find the difference quotient of the function below. \[ f(x)=4 x^{2}+6 x+6 \] 8. Find the difference quotient, \( \frac{f(m+h)-f(m)}{(m+h)-m} \), of the function below. In your answer, leave your denominator in factored form. \[ f(m)=-\frac{3}{(-2 m+1)} \]

To find the difference quotient for each function, you use the formula \[ \frac{f(x+h)-f(x)}{h}. \] Calculate \( f(x+h) \) and then substitute and simplify the expression. For the function \( f(b) = -7b - 7 \): 1. \( f(b+h) = -7(b+h) - 7 = -7b - 7h - 7 \) Then, the difference quotient is \[ \frac{(-7b - 7h - 7) - (-7b - 7)}{h} = \frac{-7h}{h} = -7. \] For \( f(k) = 4k - 5 \): 2. \( f(k+h) = 4(k+h) - 5 = 4k + 4h - 5 \) The difference quotient is \[ \frac{(4k + 4h - 5) - (4k - 5)}{h} = \frac{4h}{h} = 4. \] For \( f(c) = -2c - 9 \): 4. \( f(c+h) = -2(c+h) - 9 = -2c - 2h - 9 \) The difference quotient becomes \[ \frac{(-2c - 2h - 9) - (-2c - 9)}{h} = \frac{-2h}{h} = -2. \] For \( f(x) = 4x^2 + 6x + 6 \): 6. \( f(x+h) = 4(x+h)^2 + 6(x+h) + 6 = 4(x^2 + 2xh + h^2) + 6x + 6h + 6 \) Thus, the difference quotient is \[ \frac{(4(x^2 + 2xh + h^2) + 6x + 6h + 6) - (4x^2 + 6x + 6)}{h} = \frac{8xh + 4h^2 + 6h}{h} = 8x + 4h + 6. \] For \( f(m) = -\frac{3}{-2m + 1} \): 8. First, compute \( f(m+h) = -\frac{3}{-2(m+h) + 1} = -\frac{3}{-2m - 2h + 1} \). The difference quotient becomes \[ \frac{-\frac{3}{-2m - 2h + 1} + \frac{3}{-2m + 1}}{h} = \frac{\frac{3(-2m + 1) - 3(-2m - 2h + 1)}{(-2m + 1)(-2m - 2h + 1)}}{h} = \frac{6h}{h(-2m + 1)(-2m - 2h + 1)} = \frac{6}{(-2m + 1)(-2m - 2h + 1)}. \] And there you have it—the difference quotients for each function!